m, m and m

6267 /www/nrich/html/content/id/6267/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><p>There are several sets of five positive whole numbers with the following properties:</p> <ul> <li>Mean = 4</li> <li>Median = 3</li> <li>Mode = 3</li> </ul> <div>Can you find <span style="font-weight: bold;">all</span> the different sets of five positive whole numbers that satisfy these conditions?</div> <div>Can you convince us you have found them all?</div> <p>If I also tell you that the range is 10, can you identify my numbers?<br></br> <br></br> <br></br> <span style="font-weight: bold;">Final Challenge</span><br></br> <br></br> How many sets of five positive whole numbers are there with the following property?<br></br> </p> <div style="text-align: center;">Mean = Median = Mode = Range = a single digit number</div> <br></br> <p><a href="http://nrich.maths.org/8012">Click here for a poster of this problem</a>.</p></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p class="editorial">Many of you have managed to work out some of the different sets of positive numbers which statisfy the threeconditions: mean = 4, median = 3, and mode = 3.<br></br> Congratulation to Grace and Lydia from Hethersett High School who successfully found all the possible sets and convinced us that they had found them all with their excellent demonstration. Here is their answer:</p> <p>The problem tells us that there are 5 positive whole numbers that have a mean of 4, a median of 3, and a mode of 3. What could these numbers be?<br></br> <br></br> First of all, because the median is 3, we know that if we write the numbers in numerical order, the middle number has to be 3.<br></br> <br></br> Because the mode is 3, there must be at least one other 3, and since we are writing this in numerical order, a 3 must at least occupy a place on one side of the middle 3.<br></br> <br></br> 1st number: Has to be equal to, or less than 3<br></br> 2nd number: Has to be equal to, or less than 3<br></br> 3rd number: Has to be 3<br></br> 4th number: Has to be equal to, or greater than 3<br></br> 5th number: Has to be greater than 3 (if the largest number was 3, the mean could not be 4)<br></br> <br></br> Since the mean is 4, we know that all the numbers must add up to 20 (20 divided by the number of numbers (5) is 4). We can guarantee two 3s in the group, so the remaining numbers must total 14.<br></br> <br></br> Now we need to make a list of all the possible groups whose starting number is less than 3, and (apart from 3) no numbers are repeated twice. The only time that this repetition would work, would be with three 3s, and two of another number, but no whole number multiplied by 2 equals 11 (20 minus three 3s).<br></br> <br></br> Here is our list:<br></br> 1, 2, 11<br></br> 1, 3, 10<br></br> 1, 4, 9<br></br> 1, 5, 8<br></br> 1, 6, 7 (1, 7, 6 = 1, 6, 7 because ultimately the numbers will be written numerically).<br></br> 2, 3, 9<br></br> 2, 4, 8<br></br> 2, 5, 7<br></br> 3, 3, 8<br></br> 3, 4, 7<br></br> 3, 5, 6<br></br> <br></br> Written with the other numbers, the full sets are:<br></br> 1, 2, 3, 3, 11<br></br> 1, 3, 3, 3, 10<br></br> 1, 3, 3, 4, 9<br></br> 1, 3, 3, 5, 8<br></br> 1, 3, 3, 6, 7<br></br> 2, 3, 3, 3, 9<br></br> 2, 3, 3, 4, 8<br></br> 2, 3, 3, 5, 7<br></br> 3, 3, 3, 3, 8<br></br> 3, 3, 3, 4, 7<br></br> 3, 3, 3, 5, 6<br></br> <br></br> There are eleven sets. We have found all possible sets because we wrote them in ascending order, with the smallest possible number in first position, then the smallest possible in second position and the remainder in the last position. After that we continue to write the smallest number in first position till we had exhausted all possibilities for the other two numbers, then moved on to the second smallest number in first position, and so on. Finally we deleted all repeating combinations, and all the final sets satisfy all conditions.</p> <p><span class="editorial">Jacob (Brewood Middle School) also found the possible set. But he found 17 possibilities in total because he included 0 in his answer. Some of you also made the same mistake of including 0 as a positive number.</span></p> <p>0,1,3,3,13<br></br> 0,2,3,3,12<br></br> 0,3,3,3,11<br></br> 0,3,3,4,10<br></br> 0,3,3,5,9<br></br> 0,3,3,6,8<br></br> 1,2,3,3,11<br></br> 1,3,3,3,10<br></br> 1,3,3,4,9<br></br> 1,3,3,5,8<br></br> 1,3,3,6,7<br></br> 2,3,3,3,9<br></br> 2,3,3,4,8<br></br> 2,3,3,5,7<br></br> 3,3,3,3,8<br></br> 3,3,3,4,7<br></br> 3,3,3,5,6<br></br> <br></br> <span class="editorial">When the range of 10 was added to the three original conditions, Ha-young (Wesley College), Chris and Matt (Staunton and Corse), Bethaney and Kieran (Staunton and Corse), James (Longthorpe Primary School), and Grace and Lydia (Hethersett High School) all sent in the correct solution.</span><br></br> <br></br> <span class="editorial">Grace and Lydia followed their answer given above and concluded that:</span><br></br> <br></br> If the range is 10 the numbers must be 1, 2, 3, 3, 11, because this is the only group with a range of 10.</p> <p class="editorial">Other good reasoning was sent by James (Longthorpe Primary School):</p> <p>There must be five numbers so:<br></br> [?] [?] [?] [?] [?]<br></br> The median is three so:<br></br> [?] [?] [3] [?] [?]<br></br> The mode is also three so there must be at least two 3's so:<br></br> [?] [3] [3] [?] [?]<br></br> The mean is four so because there are five numbers, all the numbers, when added together, must make 20:<br></br> [?]+[3]+[3]+[?]+[?]= 20<br></br> <br></br> The range must be 10:<br></br> First let's try 3 - 13;<br></br> [3] [?] [?] [?] [13]<br></br> <br></br> This will not work. Because 3+13=16, we need 4 more to add up to 20.<br></br> We need two 3's, which makes 19.<br></br> This leaves only 1 and 2 empty spaces, which means 0.5 per space.<br></br> This cannot happen because each space has to be a whole number.<br></br> <br></br> Now let's try 1 - 11.<br></br> There must be two 3's so, 1+11+6=18 or [1] [?] [3] [3] [11].<br></br> There is 2more to add up to 20 so<br></br> [1] + [2] + [3] + [3] + [11] = 20.<br></br> </p> <p><span class="editorial">With the new set of conditions: mean = 31, median = 33, mode = 34, and range = 8, all of you, who have attempted to solve the problem have found the right answer. Congratulation to Daniel (Haberdashers' Aske's Boys' School), Rachel (West Moors Middle), Ha-young (Wesley College), Chris and Matt (Staunton and Corse), Bethaney and Kieran (Staunton and Corse), James (Longthorpe Primary School), and Grace and Lydia (Hethersett High School). This is a very clear explanation from Grace and Lydia:</span></p> <div>The second question tells us that another set of 5 positive whole numbers have a mean of 31, a median of 33, a mode of 34, and a range of 8.</div> <br></br> <div>Once again, the middle number has to be 33, and both places numerically above it must contain a 34, since at least two 34s are needed, and there are only 2 spaces a 34 could go into without affecting the median.</div> <div>1st number: Has to be less than 33.</div> <div>2nd number: Has to be less than 33.</div> <div>3rd number: Has to be 33.</div> <div>4th number: Has to be 34.</div> <div>5th number: Has to be 34.</div> <br></br> <div>Because the largest number is now 34, and the range is 8, the smallest number has to be 26 (34 - 8 = 26).</div> <br></br> <div>To find the final number, as the mean is 31, we must multiply 31 by 5 to get the total of all 5 numbers. If we add up the 4 numbers we have found, and then subtract this sum from 31 x 5, we will get the final number:</div> <div>155 - 127 = 28.</div> <br></br> <div>So our set is 26, 28, 33, 34, 34</div> <br></br> <div>This is the only set of numbers there is because the largest three numbers have to be where they are to satisfy specifications, and this fixes the first number through the range. This leaves only one number to be decided, and if that number is not 28, it will not satisfy the mean of 31. Therefore only one set of numbers meets the requirements.</div> <br></br> <p class="editorial">Jacob also tackled the problem with the same method but different presentation:</p> <div>There must be five numbers so:</div> <div>[?] [?] [?] [?] [?]</div> <div>The median is 33 so:</div> <div>[?] [?] [33] [?] [?]</div> <div>The mode is also 34, so there must be at least two 34's, so:</div> <div>[?] [?] [33] [34] [34]</div> <div>The mean is 31 so because there are five numbers, all the numbers, when added together, must make 155 (31x5):</div> <div>[?]+[?]+[33]+[34]+[34]= 155</div> <div>Now the numbers must add up to 155 and there also must be two 34's. Also, the range must be eight and our highest number is 34 so, 34-8=26, so our lowest number must be 26:</div> <div>[26] [?] [33] [34] [34]</div> <div>All the numbers we have so far add up to 127.</div> <div>Because155-127=28, 28 must be our final number.</div> <div>[26] [28] [33] [34] [34]</div> <div>There must only be one answer to this when there is a range as well.</div> <p class="editorial"> </p> <p class="editorial">Ha-young had a convincing reason why that is the only set we can find:</p> <p>I am convinced that the set 26, 28, 33, 34, 34 is the only set we can find because 33 needs to stay in the middle and at least two 34 needs to stay, so we cannot have higher number. If we do, the median will change. The range also needs to be 8. Therefore the minimum number has to be 26.<br></br> <br></br> <br></br> <span class="editorial">Students from Kellett School in Hong Kong had a go at the <strong>Final Challenge</strong>:</span></p> <p><br></br> Sean found the numbers 1, 2, 2, 2, 3 and 2, 5, 5, 6, 7 and 4, 8, 8, 8, 12 fit the criteria.<br></br> Sebastian found the numbers 2, 4, 4, 4, 6 and 3, 6, 6, 6, 9 fit the criteria.<br></br> Oliver found that 3, 4, 5, 5, 8 works too.</p> <p> </p> <p><span class="editorial">Students from Woodcot Primary School in Gosport found almost all the solutions to the Final Challenge:</span><br></br> <br></br> 1, 2, 2, 2, 3<br></br> <br></br> 2, 4, 4, 4, 6<br></br> <br></br> 3, 4, 5, 5, 8<br></br> 2, 5, 5, 6, 7<br></br> <br></br> 3, 6, 6, 6, 9<br></br> <br></br> 3, 7, 7, 8, 10<br></br> <br></br> 3, 8, 8, 10, 11<br></br> 4, 8, 8, 8, 12<br></br> 5, 6, 8, 8, 13</p> <p><br></br> 5, 8, 9, 9, 14</p> <br></br> <p><span class="editorial">George Simpson,</span> <span class="editorial">from St. Gabriel's High School in Bury, found one more solution:</span></p> <p>4, 9, 9, 10, 13<br></br> <br></br> <span class="editorial">Mrs O’Clee class at</span> <span class="editorial">Berkhamsted School</span> <span class="editorial">completed the set with:</span></p> <p>4, 6, 7, 7, 11</p> <br></br> <span class="editorial">Well done to you all.</span><br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <h3>Why do this problem?</h3> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6267&part=index">This problem</a> offers the students an opportunity to consolidate what they are expected to know about mean, mode and median whilst also challenging them to work systematically, and justify their reasoning.<br></br> <h3><span style="font-weight: bold;">Possible approach</span></h3> Start by writing five numbers on the board: 5, 3, 6, 3, 3 and ask for the mean, median and mode of this set. Resolve any disagreements.<br></br> "OK so this is too easy for you - what if I had told you that the mean, mode and median of five positive whole numbers were<br></br> <div style="margin-left: 40px;">mean: 4</div> <div style="margin-left: 40px;">mode: 3</div> <div style="margin-left: 40px;">median: 3</div> <div>Would you have been able to tell me the five numbers?"</div> <div>"Are there any other sets of five numbers that fit these conditions?"</div> <div>Collect a few suggestions and then ask:</div> <div>"There seem to be quite a few - can you find some more?"<br></br> "Can you find them all?"</div> <br></br> <div>Allow some time for individual work and then ask:</div> <div>"Can you convince yourself you have them all?"</div> <div>"Can you convince a friend?"</div> <div>"Can you convince the rest of the class?"</div> <br></br> <div>It would be useful to give some time for students to organise their set of solutions for sharing with the whole class.</div> <div>Invite early finishers to list their set on the board in a way that makes their system explicit.</div> <div>You may end up with several different orderings.</div> <div>You could either:</div> <div>invite each pair of writers to explain their logic to the class, OR</div> <div>invite the class to work out what the reasoning is behind each ordering and ask the writers to confirm, OR</div> <div>ask students to just list the first few sets of numbers and ask the class to predict which sets will follow.</div> <br></br> <div>"If I had also told you that the range of these five numbers was 10, how many solutions would there have been?"</div> <div>Could you add one number to these five and still have the same mean, mode, median and range?</div> <div>Could you add two numbers?</div> <div>Could you add three numbers?....</div> <br></br> Finish the activity by asking the students to make up a similar question including mean, mode, median and range, for their partner. Can they find a question which has a unique solution? <h3>Key questions</h3> Which piece of information is the most useful to start with?<br></br> What process allows you to be confident that you will have found all the results by the end?<br></br> <h3>Possible extension</h3> <div>"You should have found quite a few solutions...</div> <div>Could we have chosen a mean, mode and median for which there was only one solution?</div> <div>Could we have chosen a mean, mode and median for which there were no solutions?</div> <div>Would your answers be different if you were allowed to include negative numbers?"<br></br> <br></br> The Final Challenge from the <a href="/6267">problem</a> also makes a good follow-up question:<br></br> <br></br> How many sets of five positive whole numbers are there with the following property?</div> <div style="text-align: center;">Mean = Median = Mode = Range = a single digit number</div> <br></br> <h3>Possible support</h3> Students who find it difficult to work systematically may be helped by recording each solution on a separate slip of paper and rearranging them into 'families'. If you are using an interactive white board you could model this by recording solutions as they are produced and rearranging them.<br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> Try arranging all sets of numbers in order, smallest to biggest. Which of the five positions are the easiest to fill in? <br></br> If you can think of lots of possible values for one space, try to figure out the highest and the lowest it could be, and whether there are any other conditions...<br></br> <br></br> Which spaces can you fill straight away?<br></br> <br></br> <a href="/content/id/6267/balance.jpg"></a><a href="http://nrich.maths.org/content/id/4725/balancer.swf">This</a> might help.<br></br> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> Here are two lists produced systematically in different ways:<br></br> 1 2 3 3 11<br></br> <br></br> 1 3 3 3 10<br></br> 1 3 3 4 9<br></br> 1 3 3 5 8<br></br> 1 3 3 6 7<br></br> <br></br> 2 3 3 3 9<br></br> 2 3 3 4 8<br></br> 2 3 3 5 7<br></br> <br></br> 3 3 3 3 8<br></br> 3 3 3 4 7<br></br> 3 3 3 5 6<br></br> <br></br> *******<br></br> <br></br> 3 3 3 3 8<br></br> <br></br> 3 3 3 4 7<br></br> 3 3 3 5 6<br></br> <br></br> 1 3 3 3 10<br></br> 2 3 3 3 9<br></br> <br></br> 1 3 3 4 9<br></br> 1 3 3 5 8<br></br> 1 3 3 6 7<br></br> 2 3 3 4 8<br></br> 2 3 3 5 7<br></br> <br></br> 1 2 3 3 11<br></br> <br></br> *******<br></br> <br></br> 1, 2, 3, 3, 11 has a range of 10<br></br> <br></br> <br></br> <br></br> Final Challenge:<br></br> <br></br> 0 0 0 0 0<br></br> <br></br> 1 2 2 2 3<br></br> <br></br> 2 4 4 4 6<br></br> <br></br> 3 4 5 5 8<br></br> 2 5 5 6 7<br></br> <br></br> 3 6 6 6 9<br></br> <br></br> 3 7 7 8 10<br></br> 4 6 7 7 11<br></br> <br></br> 3 8 8 10 11<br></br> 4 8 8 8 12<br></br> 5 6 8 8 13<br></br> <br></br> 4 9 9 10 13<br></br> 5 8 9 9 14<br></br></mdoxml> 2 3 0 0 1 0 0 m, m and m If you are given the mean, median and mode of five positive whole numbers, can you find the numbers? Handling, Processing and Representing Data Mode Handling, Processing and Representing Data Median Handling, Processing and Representing Data Mean Handling, Processing and Representing Data Averages Using, Applying and Reasoning about Mathematics Working systematically Information and Communications Technology smartphone Secondary Mapping Document Statistics – processing and representing Secondary Mapping Document DisplayCabinet Secondary processes Working Systematically